Atoms & Nuclei

JEE Main level, single correct, medium

237 of 239

Question 237 of 239

The binding energy per nucleon of deuteron (12H)\left({ }_1^2 \mathrm{H}\right) and helium nucleus (24He)\left({ }_2^4 \mathrm{He}\right) is 1.1MeV1.1 \mathrm{MeV} and 7MeV7 \mathrm{MeV} respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is

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Answer: (C) 23.6MeV23.6 \mathrm{MeV}

Energy released == total binding energy of product −- total binding energy of reactants ⇒28−(2×2.2)=28−4.4=236MeV\Rightarrow 28-(2 \times 2.2)=28-4.4=236 \mathrm{MeV}

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